Pearson Edexcel IGCSE Chemistry: Physical Chemistry — Practice Questions (4CH1)

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available from your board.

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Questions

1. A student burns 0.46 g of ethanol, C₂H₅OH, to heat 100 g of water. The temperature of the water rises by 25.0 °C.

(a) Calculate the heat energy transferred to the water, in J. (c = 4.2 J/g/°C) [1]

(b) Calculate the molar enthalpy change of combustion of ethanol, in kJ/mol. (Mᵣ of ethanol = 46) [2]

2. Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide. Explain how a catalyst increases the rate of a reaction. [2]

3. Explain, in terms of particles, why powdered calcium carbonate reacts faster with dilute hydrochloric acid than the same mass of large lumps. [2]

4. Ammonia is made in a reversible reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The forward reaction is exothermic.

(a) State what is meant by a dynamic equilibrium. [2]

(b) State and explain the effect on the yield of ammonia of increasing the temperature. [2]

5. Use the bond energies below to calculate the enthalpy change for H₂ + Cl₂ → 2HCl.

Bond energies in kJ/mol: H–H 436, Cl–Cl 242, H–Cl 431 [3]

Answers

1. (a) Q = mcΔT = 100 × 4.2 × 25.0 = 10 500 J [1]. (b) Moles of ethanol = 0.46 ÷ 46 = 0.0100 mol [1]; ΔH = −10.5 kJ ÷ 0.0100 mol = −1050 kJ/mol [1] (negative because the reaction is exothermic).

Mark-scheme insight (Pearson Edexcel 4CH1 June 2024 mark scheme, Paper 1CR, Question 6(b)(ii)–(iii)): the energy calculation uses the mass of water heated × 4.2 × the temperature rise; the next part then has separate marks for the energy in kJ, the moles of fuel, the division and a final answer with a negative sign. Try the real question next: Pearson Edexcel International GCSE Chemistry 4CH1, June 2024, Paper 1CR, Question 6.

Source for the mark-scheme insights on this page: Pearson Edexcel International GCSE Chemistry 4CH1 Paper 1CR mark scheme (Summer 2024), paraphrased.

2. A catalyst provides an alternative pathway (route) for the reaction [1] with a lower activation energy, so more collisions have enough energy to react [1].

Mark-scheme insight (Pearson Edexcel 4CH1 June 2024 mark scheme, Paper 1CR, Question 10(a)): the two marks are for an alternative pathway (route) and a lower activation energy. Our tip: saying only that a catalyst “speeds up the reaction” does not explain how it works. Try the real question next: Pearson Edexcel International GCSE Chemistry 4CH1, June 2024, Paper 1CR, Question 10.

3. Powder has a larger surface area (to volume ratio), so more particles of calcium carbonate are exposed to the acid [1], giving more frequent collisions between acid particles and calcium carbonate particles [1].

4. (a) The forward and backward reactions happen at the same rate [1], so the concentrations of reactants and products stay constant (in a closed system) [1]. (b) The yield decreases [1], because the equilibrium shifts in the endothermic (backward) direction to oppose the rise in temperature [1].

5. Bonds broken: H–H + Cl–Cl = 436 + 242 = 678 kJ/mol [1]. Bonds made: 2 × H–Cl = 862 kJ/mol [1]. ΔH = 678 − 862 = −184 kJ/mol [1].

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Where marks are usually lost

• Using the mass of the fuel instead of the mass of water in Q = mcΔT.
• Leaving out the negative sign for an exothermic enthalpy change.
• Saying a catalyst “gives particles more energy” instead of lowering the activation energy.
• Describing equilibrium as “the reaction has stopped”.